Certainly! Let's go through the problem step by step.
We are given a polynomial [tex]\( p(x) = x^2 - 2x + 3 \)[/tex] and we need to find a new polynomial whose roots are [tex]\(\alpha + 2\)[/tex] and [tex]\(\beta + 2\)[/tex], where [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the roots of the given polynomial.
1. Find the roots ([tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex]) of the polynomial [tex]\( p(x) = x^2 - 2x + 3 \)[/tex]:
- The roots of the polynomial can be determined using the quadratic formula:
[tex]\[
\alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
- For [tex]\( p(x) = x^2 - 2x + 3 \)[/tex], we have [tex]\(a = 1\)[/tex], [tex]\(b = -2\)[/tex], and [tex]\(c = 3\)[/tex].
[tex]\[
\alpha, \beta = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 12}}{2} = \frac{2 \pm \sqrt{-8}}{2} = \frac{2 \pm 2i\sqrt{2}}{2} = 1 \pm i\sqrt{2}
\][/tex]
- Thus, the roots are:
[tex]\[
\alpha = 1 - i\sqrt{2}, \quad \beta = 1 + i\sqrt{2}
\][/tex]
2. Determine the new roots [tex]\(\alpha + 2\)[/tex] and [tex]\(\beta + 2\)[/tex]:
- If [tex]\(\alpha\)[/tex] is a root, [tex]\(\alpha + 2\)[/tex] will be its shifted version:
[tex]\[
\alpha + 2 = (1 - i\sqrt{2}) + 2 = 3 - i\sqrt{2}
\][/tex]
- Similarly:
[tex]\[
\beta + 2 = (1 + i\sqrt{2}) + 2 = 3 + i\sqrt{2}
\][/tex]
- Therefore, the new roots are:
[tex]\[
3 - i\sqrt{2}, \quad 3 + i\sqrt{2}
\][/tex]
3. Form the new polynomial with these new roots:
- A polynomial with roots [tex]\(r_1\)[/tex] and [tex]\(r_2\)[/tex] can be expressed as:
[tex]\[
(x - r_1)(x - r_2)
\][/tex]
- Substituting the new roots:
[tex]\[
(x - (3 - i\sqrt{2}))(x - (3 + i\sqrt{2}))
\][/tex]
- Simplify this expression:
[tex]\[
(x - 3 + i\sqrt{2})(x - 3 - i\sqrt{2})
\][/tex]
4. Expand the polynomial:
- Use the difference of squares formula [tex]\((a - b)(a + b) = a^2 - b^2\)[/tex]:
[tex]\[
(x - 3 + i\sqrt{2})(x - 3 - i\sqrt{2}) = (x - 3)^2 - (i\sqrt{2})^2
\][/tex]
- Calculate each term:
[tex]\[
(x - 3)^2 = x^2 - 6x + 9
\][/tex]
[tex]\[
(i\sqrt{2})^2 = -2
\][/tex]
- Combine these results:
[tex]\[
x^2 - 6x + 9 - (-2) = x^2 - 6x + 9 + 2 = x^2 - 6x + 11
\][/tex]
Therefore, the polynomial whose roots are [tex]\(\alpha + 2\)[/tex] and [tex]\(\beta + 2\)[/tex] is:
[tex]\[
\boxed{x^2 - 6x + 11}
\][/tex]
